Integrating trigonometric function

Question

How do you integrate: $$\int_{0}^{1}\frac{\sin(x)}{\sin(1-x)+\sin(x)}dx$$

The answer that you must get is $\frac{1}{2}.$

Answer 1

Let $I=\int_0^1\frac{\sin x}{\sin x+\sin (1-x)}\,dx~~~~~~~~~~~~~~~~(1)$

Take $x=1-z$. Then $I=\int_0^1\frac{\sin (1-z)}{\sin (1-z)+\sin (z)}\,dz$. Or we can write it as $$I=\int_0^1\frac{\sin (1-x)}{\sin x+\sin (1-x)}\,dx~~~~~~~~~(2)$$

Add $(1)$ and $(2)$, we have

$$2I=\int_0^1 1\,dx \implies I=\frac{1}{2}.$$